Answer
Divergent
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{3-\cos n}{n^{2 / 3}-2}$$
Since $ n^{2 / 3}-2> 0$ for $n\geq 3$, then
$$
\frac{3-\cos n}{n^{2 / 3}-2}>\frac{1}{n^{2 / 3}-2}>\frac{1}{n^{2 / 3}},\ \ n \geq 3
$$
Since $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2 / 3}}$ is divergent ($p-$series $p<1$), then by the comparison test, $\displaystyle\sum_{n=1}^{\infty} \frac{3-\cos n}{n^{2 / 3}-2}$ also diverges.